Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 24 September 2014

Lecture 23: Quotient groups

Let G be a group and let H be a subgroup of G. The set of cosets of H in G is GH= {gH|gG}, where gH={gh|hH} for gG.

A normal subgroup of G is a subgroup K of G such that if gG and kK then gkg-1K.

Let G be a group and let H be a subgroup of G. Then GH with GH×GH GH (g1H,g2H) g1g2H is a well defined group if and only if H is a normal subgroup of G.

If G = { , , , , , } =S3and H = { , } the cosets in GH are ·H= ·H= { , } , ·H= ·H= { , } , ·H= ·H= { , }

If we try to define GH×GH GH (g1H,g2H) g1g2H (*) then (·H) (·H)= ·H= { , } = (·H) (·H)= ·H= { , } gives { , } 2 = { , } and { , } 2 = { , } . In other words, for this H, (*) is not a function! The theorem tells us that this is "because" H is not normal: H, but · · ()-1= · · = = is not in H.

Proof of the theorem.

:
Assume H is a normal subgroup.
To show:
(a) GH×GH GH (g1H,g2H) g1g2H is a function.
(b) Using (g1H)(g2H)=g1g2H as product:
(ba) If g1H,g2H,g3HGH then (g1Hg2H) g3H=g1H (g2Hg3H).
(bb) There exists eHGH such that if gHGH then (eH)(gH)=gH and (gH)(eH)=gH.
(bc) If gHGH then there exists xHGH such that (xH)(gH)=eH and (gH)(xH)=eH.
(a) To show: If (g1H,g2H)=(g1H,g2H) then g1g2H=g1g2H.
Assume (g1H,g2H)=(g1,g2H).
Then g1H=g1H and g2H=g2.
To show: g1g2H=g1g2H.
To show: g1g2g1g2H, since the cosets partition G.
We know g1g1H and g2g2H.
So there exist h1,h2H such that g1=g1h1 and g2=g2h2.
To show: g1g2g1g2H. g1g2 = g1h1 g2h2 = g1g2 (g2)-1 h1g2h2 = g1g2 ((g2)-1h1g2) h2g1g2H since H is a normal subgroup of G.
(b) Use (g1H)(g2H)=g1g2H as product in GH.
(ba) To show: If g1H,g2H,g3HGH then (g1Hg2H)g3H=g1H(g2Hg3H).
Assume g1H,g2H,g3HGH.
To show: (g1Hg2H)g3H=g1H(g2Hg3H).
Then (g1Hg2H)g3H= g1g2H·g3H= (g1g2)g3H and g1H(g2Hg3H)= g1H·g2g3H= g1(g2g3)H and (g1g2)g3=g1(g2g3) since G is associative.
So (g1Hg2H)g3H=g1H(g2Hg3H).
(bb) To show: There exist eHGH such that if gHGH then (eH)(gH)=gH and (gH)(eH)=gH. Let eH=H.
To show: if gHGH then (eH)(gH)=gH and (gH)(eH)=gH.
Assume gHGH.
To show:
(bba) eH·gH=gH.
(bbb) gH·eH=gH.
(bba) eH·gH=1·H·g·H =1(g)H=gH, since 1 is an identity for G.
(bbb) gHeH=gH1·H= (g1)·H=gH.
(bc) To show: If gHGH then there exists xHGH such that (gH)(xH)=eH and xHgH=H.
Assume gHGH.
To show: There exists xHGH such that gHxH=Hand xHgH=H. Let xH=g-1H.
To show:
(bca) gHxH=H.
(bcb) xHgH=H.
(bca) gHxH=gHg-1H =gg-1H=1·H=H.
(bcb) xHgH= g-1HgH= (gg-1)H= 1·H=H.
This completed the proof that if H is a normal subgroup of G then GH with (g1H)(g2H)=g1g2H is a group.

Notes and References

These are a typed copy of Lecture 23 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on September 14, 2011.

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